Integrand size = 14, antiderivative size = 159 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {2 b x}{3 c}+\frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {b \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}} \]
-2/3*b*x/c+1/3*x^3*(a+b*arctan(c*x^2))+1/6*b*arctan(-1+x*2^(1/2)*c^(1/2))/ c^(3/2)*2^(1/2)+1/6*b*arctan(1+x*2^(1/2)*c^(1/2))/c^(3/2)*2^(1/2)-1/12*b*l n(1+c*x^2-x*2^(1/2)*c^(1/2))/c^(3/2)*2^(1/2)+1/12*b*ln(1+c*x^2+x*2^(1/2)*c ^(1/2))/c^(3/2)*2^(1/2)
Time = 0.06 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.11 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {2 b x}{3 c}+\frac {a x^3}{3}+\frac {1}{3} b x^3 \arctan \left (c x^2\right )+\frac {b \arctan \left (\frac {-\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \arctan \left (\frac {\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}} \]
(-2*b*x)/(3*c) + (a*x^3)/3 + (b*x^3*ArcTan[c*x^2])/3 + (b*ArcTan[(-Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(3*Sqrt[2]*c^(3/2)) + (b*ArcTan[(Sqrt[2] + 2*Sqr t[c]*x)/Sqrt[2]])/(3*Sqrt[2]*c^(3/2)) - (b*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x ^2])/(6*Sqrt[2]*c^(3/2)) + (b*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(6*Sqrt[ 2]*c^(3/2))
Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5361, 843, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \int \frac {x^4}{c^2 x^4+1}dx\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\int \frac {1}{c^2 x^4+1}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \int \frac {c x^2+1}{c^2 x^4+1}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} \sqrt {c} x\right )^2-1}d\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{-\left (\sqrt {2} \sqrt {c} x+1\right )^2-1}d\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 \sqrt {2} c}+\frac {\int \frac {\sqrt {2} \sqrt {c} x+1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}-\frac {\log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}\right )}{c^2}\right )\) |
(x^3*(a + b*ArcTan[c*x^2]))/3 - (2*b*c*(x/c^2 - ((-(ArcTan[1 - Sqrt[2]*Sqr t[c]*x]/(Sqrt[2]*Sqrt[c])) + ArcTan[1 + Sqrt[2]*Sqrt[c]*x]/(Sqrt[2]*Sqrt[c ]))/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2]/(Sqrt[2]*Sqrt[c]) + Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2]/(2*Sqrt[2]*Sqrt[c]))/2)/c^2))/3
3.1.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Time = 0.54 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(\frac {x^{3} a}{3}+b \left (\frac {x^{3} \arctan \left (c \,x^{2}\right )}{3}-\frac {2 c \left (\frac {x}{c^{2}}-\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2}}\right )}{3}\right )\) | \(118\) |
| parts | \(\frac {x^{3} a}{3}+b \left (\frac {x^{3} \arctan \left (c \,x^{2}\right )}{3}-\frac {2 c \left (\frac {x}{c^{2}}-\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2}}\right )}{3}\right )\) | \(118\) |
1/3*x^3*a+b*(1/3*x^3*arctan(c*x^2)-2/3*c*(1/c^2*x-1/8/c^2*(1/c^2)^(1/4)*2^ (1/2)*(ln((x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2-(1/c^2)^(1/4)*x *2^(1/2)+(1/c^2)^(1/2)))+2*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)+2*arctan(2^(1 /2)/(1/c^2)^(1/4)*x-1))))
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.96 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {2 \, b c x^{3} \arctan \left (c x^{2}\right ) + 2 \, a c x^{3} + c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b x + c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}}\right ) + i \, c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b x + i \, c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}}\right ) - i \, c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b x - i \, c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}}\right ) - c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b x - c \left (-\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}}\right ) - 4 \, b x}{6 \, c} \]
1/6*(2*b*c*x^3*arctan(c*x^2) + 2*a*c*x^3 + c*(-b^4/c^6)^(1/4)*log(b*x + c* (-b^4/c^6)^(1/4)) + I*c*(-b^4/c^6)^(1/4)*log(b*x + I*c*(-b^4/c^6)^(1/4)) - I*c*(-b^4/c^6)^(1/4)*log(b*x - I*c*(-b^4/c^6)^(1/4)) - c*(-b^4/c^6)^(1/4) *log(b*x - c*(-b^4/c^6)^(1/4)) - 4*b*x)/c
Time = 7.49 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {atan}{\left (c x^{2} \right )}}{3} + \frac {b \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (c x^{2} \right )}}{3} - \frac {2 b x}{3 c} - \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x - \sqrt [4]{- \frac {1}{c^{2}}} \right )}}{3 c} + \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{6 c} + \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {x}{\sqrt [4]{- \frac {1}{c^{2}}}} \right )}}{3 c} & \text {for}\: c \neq 0 \\\frac {a x^{3}}{3} & \text {otherwise} \end {cases} \]
Piecewise((a*x**3/3 + b*x**3*atan(c*x**2)/3 + b*(-1/c**2)**(3/4)*atan(c*x* *2)/3 - 2*b*x/(3*c) - b*(-1/c**2)**(1/4)*log(x - (-1/c**2)**(1/4))/(3*c) + b*(-1/c**2)**(1/4)*log(x**2 + sqrt(-1/c**2))/(6*c) + b*(-1/c**2)**(1/4)*a tan(x/(-1/c**2)**(1/4))/(3*c), Ne(c, 0)), (a*x**3/3, True))
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \arctan \left (c x^{2}\right ) - c {\left (\frac {8 \, x}{c^{2}} - \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}}{c^{2}}\right )}\right )} b \]
1/3*a*x^3 + 1/12*(4*x^3*arctan(c*x^2) - c*(8*x/c^2 - (2*sqrt(2)*arctan(1/2 *sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/sqrt(c) + 2*sqrt(2)*arctan(1/2 *sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/sqrt(c) + sqrt(2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/sqrt(c) - sqrt(2)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1 )/sqrt(c))/c^2))*b
Time = 0.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{12} \, b c^{5} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}}\right )} + \frac {b c x^{3} \arctan \left (c x^{2}\right ) + a c x^{3} - 2 \, b x}{3 \, c} \]
1/12*b*c^5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt (abs(c)))/(c^6*sqrt(abs(c))) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2) /sqrt(abs(c)))*sqrt(abs(c)))/(c^6*sqrt(abs(c))) + sqrt(2)*log(x^2 + sqrt(2 )*x/sqrt(abs(c)) + 1/abs(c))/(c^6*sqrt(abs(c))) - sqrt(2)*log(x^2 - sqrt(2 )*x/sqrt(abs(c)) + 1/abs(c))/(c^6*sqrt(abs(c)))) + 1/3*(b*c*x^3*arctan(c*x ^2) + a*c*x^3 - 2*b*x)/c
Time = 0.45 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.39 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a\,x^3}{3}+\frac {b\,x^3\,\mathrm {atan}\left (c\,x^2\right )}{3}-\frac {2\,b\,x}{3\,c}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )\,1{}\mathrm {i}}{3\,c^{3/2}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )}{3\,c^{3/2}} \]